Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<queue>
#include<utility>
#include<algorithm>

using namespace std;

vector<int> edges[123456];

int N,M,s,t;

vector<int> dijkstra(int s){
  vector<int> d(N,1e9);
  queue<pair<int,int> > que;
  que.push({s,0});
  while(!que.empty()){
    auto c=que.front();
    que.pop();
    if(d[c.first]<=c.second)continue;
    d[c.first]=c.second;
    for(auto e:edges[c.first]){
      que.push({e,c.second+1});
    }
  }
  return d;
}

int main(){
  cin>>N>>M>>s>>t;
  s--;
  t--;
  while(M--){
    int x,y;
    cin>>x>>y;
    x--;
    y--;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  auto sto=dijkstra(s);
  auto tto=dijkstra(t);
  int sp=sto[t];
  sort(begin(sto),end(sto));
  long long ans=0;
  for(auto e:tto){
    ans+=upper_bound(begin(sto),end(sto),sp-2-e)-lower_bound(begin(sto),end(sto),sp-2-e);
  }
  cout<<ans<<endl;
}